Question

Let $ f(x)= \begin{cases} \frac{x-4}{\left|x-4\right|}+a, & \quad x <\,4 \\ a+b, & \quad x=4 \\ \frac{x-4}{\left|x-4\right|}+b, & \quad x >\,4 \end{cases} $
Then $f (x)$ is continuous at $x=4$ when

• A. $a=0, b=0$
• B. $a=1, b=1$
• C. $a=-1, b=1$
• D. $a=1, b=-1$

Answer 1

Answer: (D)

We have
L.H.L $=\displaystyle\lim_{x\to4^{-}} f \left(x\right) $
$=\displaystyle\lim_{h\to0} f \left(4-h\right)$
$=\displaystyle\lim_{h\to0} \frac{4-h-4}{\left|4-h-4\right|}+a$
$=\displaystyle\lim_{h\to0} \left(-\frac{h}{h}+a\right)=a-1$
R.H.L $=\displaystyle\lim_{x\to4^{+}} f \left(x\right)$
$=\displaystyle\lim_{h\to0} f \left(4+h\right)$
$=\displaystyle\lim_{h\to0} \frac{4+h-4}{\left|4+h-4\right|}+b=b+1$
$\therefore f \left(4\right)=a+b$
Since $f \left(x\right)$ is continuous at $x = 4$,
$\displaystyle\lim_{x\to4^{-}} f \left(x\right)=f \left(4\right)=\displaystyle\lim_{x\to4^{+}} f \left(x\right)$
or $a-1=a+b=b+1$
or $b=-1$ and $a=1$